/*
Bone Collector
Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …
The bone collector had a big bag with a volume of V ,and along his trip of collecting there are a lot of bones , obviously , different bone has different value and different volume, now given the each bone’s value along his trip , can you calculate out the maximum of the total value the bone collector can get ?
Input
The first line contain a integer T , the number of cases.
Followed by T cases , each case three lines , the first line contain two integer N , V, (N <= 1000 , V <= 1000 )representing the number of bones and the volume of his bag. And the second line contain N integers representing the value of each bone. The third line contain N integers representing the volume of each bone.
Output
One integer per line representing the maximum of the total value (this number will be less than 2 31).
Sample Input
1
5 10
1 2 3 4 5
5 4 3 2 1
Sample Output
14
题意分析：所选物品不可拆，每个物品限选一次，所以是01背包问题
*/
#include <bits/stdc++.h>
#define ll long long
#define PI acos(-1)
#define M(n, m) memset(n, m, sizeof(n));
const int INF = 1e9 + 7;
const int maxn = 1e5;
using namespace std;

int T, n, weight, dp[maxn];
struct data
{
    int value;
    int weight;
} a[maxn];

int main()
{
    scanf("%d", &T);
    while(T --)
    {
        memset(dp, 0, sizeof(dp));
        scanf("%d%d", &n, &weight);
        for (int i = 0; i < n; i ++)
            scanf("%d", &a[i].value);
        for (int i = 0; i < n; i ++)
            scanf("%d", &a[i].weight);
        for (int i = 0; i < n; i ++)
            for (int j = weight; j >= a[i].weight; j --)
                dp[j] = max(dp[j], dp[j - a[i].weight] + a[i].value);
        printf("%d\n", dp[weight]);
    }
    return 0;
}
